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# 9.2: Energy and Work

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

In this section, we go beyond kinematics," and examine the underlying cause of the mtion, i.e., the dynamics."

• to identify the types of mechanical energy a body can possess
• to identify work and to determine the work done by forces on some mechanical systems
• to ascertain the work energy relationship for some mechanical systems

## Vocabulary

energy, kinetic energy, work, gravitational potential energy

Energy
Energy is a term that we hear over and over again. It is what a body possesses that allows it to do work. The more energy a body has, the more work it can do. Energy comes in many forms: chemical, electrical, nuclear, and mechanical. In this chapter we are interested only in the mechanical energy of a body. We can divide mechanical energy into two types: kinetic energy (symbol KE) and potential energy (symbol PE). The units we will use for KE and PE in this chapter will be joules (symbol J\begin{align*}J\end{align*}).
Kinetic Energy
Kinetic energy, KE, is the energy possessed by a body (of mass, m\begin{align*}m\end{align*}) that is moving with instantaneous velocity, v\begin{align*}v\end{align*}. It is expressed mathematically as follows:

KE=12mv2\begin{align*}KE=\frac{1}{2}mv^2\end{align*}

Notice that kinetic energy is always positive because the square of the velocity is also positive. If a body is not moving, then KE=0\begin{align*}KE = 0\end{align*}.

Example 1
A boy with a mass of 60.0kg\begin{align*}60.0 \;\mathrm{kg}\end{align*} runs with a velocity v=0.500m/s\begin{align*}v= -\!0.500\;\mathrm{m/s}\end{align*}. His kinetic energy is KE=12(60.0kg)(0.500ms)2=7.50J\begin{align*}KE=\frac{1}{2}(60.0 \;\mathrm{kg} )\left(-0.500\frac{m}{s} \right)^2= 7.50 \;\mathrm{J}\end{align*}
Gravitational Potential Energy
Potential energy is always associated with interactions between two or more bodies; in the case of gravitational potential energy, we consider a body with mass m\begin{align*}m\end{align*} in the gravitational pull of the Earth. GPE is the potential energy a body possesses based on its position relative to a reference level (usually the Earth’s surface). We can express GPE mathematically as:

GPE=mgh\begin{align*}GPE = mgh\end{align*}

where:

mgh=mass of crate=acceleration due to gravity=9.80 m/s2=height above a reference level\begin{align*} m &= \text{mass of crate} \\ g &= \text{acceleration due to gravity}\\ & = 9.80\ \text{m/s}^{2}\\ h &= \text{height above a reference level}\end{align*}

Notice that the higher a body is from the Earth's surface ground, the greater its GPE.

Example 2
Lift a 2.00kg\begin{align*}2.00 \;\mathrm{kg}\end{align*} box to a height of 1.50m\begin{align*}1.50 \;\mathrm{m}\end{align*} above the Earth’s surface. The gravitational potential energy of the box (relative to the Earth) is GPE=(2.00 kg)(9.80 ms2)(1.5 m)=29.4 J\begin{align*}GPE=(2.00 \ \text{kg}) \left( 9.80 \ \frac{m}{s^2} \right)(1.5 \ \text{m})=29.4 \ \text{J}\end{align*}
Example 3
A 0.500m\begin{align*}0.500 \;\mathrm{m}\end{align*} tall stool sits next to you on the Earth’s surface. When you lift a 2.00kg\begin{align*}2.00 \;\mathrm{kg}\end{align*} box to a height of 1.50m\begin{align*}1.50 \;\mathrm{m}\end{align*} above the Earth's surface, the gravitational potential energy of the box (relative to the top of the stool) is now:

GPE=(2.00 kg)(9.80 ms2)(0.500 m)=4.90 J\begin{align*}GPE=(2.00 \ \text{kg}) \left(9.80 \ \frac{m}{s^2} \right)(0.500 \ \text{m})=4.90 \ \text{J}\end{align*}

Work
Work is a term associated with our daily activities involving physical and mental stress, goals to accomplish, deadlines to meet, etc. Of course, the work we refer to as labor is different than the true definition of the word work (symbol W) we shall study in this chapter. When bodies need an applied force to move them, work is being done on the body by that applied force. The work that is done by the applied force causes the energy of the body to change. In this chapter we will study the work done by one or more forces on one or more bodies, determine the types of energy involved, and draw connections between the work done on the bodies and the energies changes in the bodies. First, however, we need to identify what we call work.

## Identifying Work

Work, W\begin{align*}W\end{align*}, is defined as the product

W=Fappd\begin{align*}W = F_{app} \cdot d_\|\end{align*}

Where Fapp\begin{align*}F_{app}\end{align*} is the force applied to a body (either a push or a pull) and d\begin{align*}d_\|\end{align*} is the displacement of the body.

Using the units of Fapp\begin{align*}F_{app}\end{align*} are newtons, N\begin{align*}N\end{align*}, and the units of “d\begin{align*}d\end{align*}” are in “meters,” m\begin{align*}m\end{align*}. So, work, W\begin{align*}W\end{align*}, is in units of “Joules," where 1Nm=1J\begin{align*}1 \;\mathrm{Nm} = 1 \;\mathrm{J}\end{align*}. Note that Nm\begin{align*}Nm\end{align*} is never used as a unit of work (or energy); rather it is reserved as the unit for "torque."

The Free Body Diagram

To determine all the work being done on a single body, we need order to clarify all the forces acting on the body. To this end, a free-body diagram of that body is usually created. The purpose of the free-body diagram is twofold:

1. to treat the single body as a “point mass” having the same amount of mass as the original body, but with a volume concentrated at one point. The reason for doing this is to circumvent any rotation that may actually occur when one or more forces are applied to the body (you can’t rotate a “point”). We are only interested here in the work that causes the body to move in one dimension.
2. to show all the forces “on” the body of interest, not the forces the body may impose on other bodies.

In creating free-body diagrams, forces acting on the point particle are always drawn as pull forces. That is, the head of the vector arrow representing each force always points away from the point mass. Think about it. Pushing or pulling on a body in the same direction, with the same amount of force, creates the same motion of the body.

## Work Done by a Single Force

Figure 8 above shows the work being done on a crate by applying a pull force of 50.0N\begin{align*}50.0\;\mathrm{N}\end{align*} created by the man. The crate is initially at rest. Figure 8(a) shows a person pulling on the rope attached to the crate causing the crate to move a distance d=0.50m\begin{align*}d= 0.50 \;\mathrm{m}\end{align*}. We will assume that friction between the bottom surface of the crate and the floor is so small that we consider it negligible, so it will not affect the motion of the crate.

Figure 8(b) is a free-body diagram of the crate. The weight of the crate is balanced by the normal force on the crate due to the surface of the floor. The only force acting on the crate is the single force due to the tension force T=50.0N\begin{align*}T = 50.0 \;\mathrm{N}\end{align*} in the rope. As the person pulls on the rope, tension in the rope does work on the crate causing it to move a distance d=0.50m\begin{align*}d = 0.50 \;\mathrm{m}\end{align*}. The amount of work done on the crate by the tension force is

W=Td=(50.0 N)(0.50 m)=25 J\begin{align*} W &= T d_{\|} \\ &= (50.0 \ \text{N})(0.50 \ \text{m}) \\ &= 25 \ \text{J} \end{align*}

This is the maximum amount of work done on the crate by the rope.

A person pulling on a wooden crate that has a mass of using rope. We will consider the rope’s mass to be so small that it can be neglected. The person pulls the crate with a force of magnitude , for a distance . There is no friction between the box and the floor. The net effect of all these forces is the movement of the box to the left. (b) A free-body diagram of the crate.

## Work Done by Several Forces

If more than one force is applied to a body, the net or total work, WT\begin{align*}W_T\end{align*}, on a body is the sum of the individual works done by the individual forces.

Example 4

The wooden crate in Figure 8 is acted upon by two forces in the vertical direction, gravity (pulling downward) and the normal force (pushing upward) from the ground.

The total work, WT\begin{align*}W_T\end{align*}, done by these two forces on the wooden crate is

WT=Wgravity+WNormal=0+0=0\begin{align*}W_T = W_{gravity} + W_{Normal} = 0 + 0 = 0\end{align*}

Because the two forces are balanced, the box does not move in the vertical direction. Hence, WT=0\begin{align*}W_T = 0\end{align*}. Note that the work done by the normal force is always zero.

## The Work—Energy Theorem

Work causes bodies to change their energy. The total work, WT\begin{align*}W_T\end{align*}, done by all forces acting on a body changes where m is the mass of the object and v is the speed of the object. This can be expressed mathematically as

WT=ΔKE\begin{align*}W_{T} = \Delta KE\end{align*}

Because the unit of KE is joules, the unit of WT\begin{align*}W_{T}\end{align*} is also joules. The person in Figure 8 increases the kinetic energy of the crate by 25J\begin{align*}25 \;\mathrm{J}\end{align*} due to the total work done on the crate.

In Figure 9, a person applies a pushing force, Fapp\begin{align*}F_{app}\end{align*} to wooden crate of mass m=2.0kg\begin{align*}m = 2.0 \;\mathrm{kg}\end{align*} crate. There is no friction between the crate and the floor due to the wheels underneath the crate. The increase in kinetic energy due to the work done by Fapp\begin{align*}F_{app}\end{align*} over a distance d=2.0m\begin{align*}d = 2.0 \;\mathrm{m}\end{align*} causes the crate to increase its initial velocity from vi=0.10m/s\begin{align*}v_i = 0.10\;\mathrm{m/s}\end{align*}, to a final velocity of vf=0.50m/s\begin{align*}v_f = 0.50\;\mathrm{m/s}\end{align*}. That is

WTor Fapp dor Fapp (2.0 m)=ΔKE=12mv2f12mv2i=12(2.0 kg)(0.50 ms)212(2.0 kg)(0.10 ms)2=0.24J\begin{align*} W_T &= \Delta KE\\ \text{or} \ F_{app} \ d &= \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2\\ \text{or} \ F_{app} \ (2.0 \ \text{m}) &= \frac{1}{2} (2.0 \ \text{kg}) \left(0.50 \ \frac{m}{s} \right)^2 - \frac{1}{2} (2.0 \ \text{kg}) \left( 0.10 \ \frac{m}{s} \right)^2 \\ & = 0.24 \, \text{J} \end{align*}

Solving for Fapp\begin{align*}F_{app}\end{align*}, we get Fapp=0.12N\begin{align*}F_{app} = 0.12 \;\mathrm{N}\end{align*}.

A person pushing a crate , initially at rest , with a constant applied force, , on frictionless ground. The direction of is indicated by the red arrow. The crate is pushed a distance, d, causing work, W, to be done on it by the . The work increases the velocity of the crate from to .

Work of Two Applied Forces on an Object

Consider the situation shown below in Figure 10. Person 1\begin{align*}1\end{align*} and dog 2\begin{align*}2\end{align*} are each pulling on a crate that has a mass of 2.0kg\begin{align*}2.0 \;\mathrm{kg}\end{align*}. Both person and the dog are pulling on a rope attached to the crate. We will consider the rope so light that we can neglect its mass. There is friction between the crate and the floor.

Person 1 and dog 2 are each pulling on a crate using ropes (with negligible mass) attached to the crate. The crate has a mass of . Person pulls with a force of magnitude . The dog pulls with a force of magnitude . The magnitude of the kinetic frictional force between the crate and the floor is .

In what follows, answer the questions pertaining to Figure 10.

Procedural Steps

1. In the box to the right of the dog in Figure 10, draw a free-body diagram for the crate, showing all the forces on it. Label each force and its magnitude.
2. The crate in Figure 10 moves a distance x=5.00m\begin{align*}x = 5.00 \;\mathrm{m}\end{align*} to the right. Calculate the work done by each applied force on the crate and the total work, WT\begin{align*}W_T\end{align*}, done on the crate, W1\begin{align*}W_1\end{align*}, W2\begin{align*}W_2\end{align*}, and Wfriction\begin{align*}W_{friction}\end{align*}
3. Determine the kinetic energy change, KE\begin{align*}\triangle KE\end{align*}, of the crate.
4. Determine the final velocity of the crate.

## Work and GPE

Work done by gravity on a body can also cause a change in the gravitational potential energy, GPE, of the body. Figure 11 shows a crate, originally at height, h, from the ground, acted upon by gravity. Because gravity is a conservative force, the work done by gravity alone on a body is independent of the path taken by the body. In this case, an equal amount of work is done on the crate by gravity alone when the crate goes from height, h, to ground level either down a frictionless hill, or down a series of steps.

The work done by gravity, Wg\begin{align*}W_g\end{align*}, is

Wg=mghcos(0)=mgh\begin{align*}W_{g}=mgh\cos(0^\circ)=mgh\end{align*}

The change in GPE for the crate going from height, h\begin{align*}h\end{align*}, to ground is

ΔGPE=GPEfGPEi=mg(0)mgh=mgh\begin{align*}\Delta GPE &= GPE_f - GPE_i \\ &= mg(0) - mgh \\ &= -mgh\end{align*}

Therefore

Wg=GPE\begin{align*}W_{g}=-GPE\end{align*}

The brown crate has two paths, and , available to it to descend from height, , to ground level . Path is a ramp with negligible friction inclined at to the horizontal (ground). Path is a series of steps. The work done by gravity, , to the crate from height to ground level is the same for each path. is “independent” of the path taken and only depends on the difference in height, . Free-body diagrams of crate for paths and are shown in the small boxes above the crates.

## Application of Work to Various Bodies

### The Mass-Pulley System

One type of system consisting of more than one body is the mass-pulley system. Figure 12 consists of two identical crates (1\begin{align*}1\end{align*} and 2\begin{align*}2\end{align*}) made up of identical masses and connected by a massless string that runs over a frictionless pulley. There is tension, T\begin{align*}T\end{align*}, in the rope. Crate 1\begin{align*}1\end{align*} is initially held at rest on a frictionless surface.

At time t=0\begin{align*}t = 0\end{align*}, crate 1\begin{align*}1\end{align*} is released and the two crates move a distance of 0.5m\begin{align*}0.5 \;\mathrm{m}\end{align*} upon which crate 1\begin{align*}1\end{align*} hits the pulley and stops. Note: A motion sensor, placed on the left side of crate, detects the instantaneous velocity of crate 1\begin{align*}1\end{align*}, can be used to determine the kinetic energy, KE, of each crate.

The mass-pulley system is made up of two identical masses (crate and crate ) connected by a massless string that runs over a pulley with negligible friction. Crate is initially held at rest on table top. The friction between crate and the table top can also be considered negligible. Both masses are . When crate is released, both crates move .

Procedural Steps

1. In the boxes to the right of the mass-pulley system in Figure 12, draw a free-body diagram for each crate, showing all the forces on it. Label each force and its magnitude.
2. Determine the work done by each force on crate 1\begin{align*}1\end{align*} and crate 2\begin{align*}2\end{align*} in Figure 12 and the total work done on both crates: W1\begin{align*}W_1\end{align*}, W2\begin{align*}W_2\end{align*}, WT\begin{align*}W_T\end{align*}.
3. Determine the change in kinetic energy,  KE\begin{align*}\triangle\ KE\end{align*}, of each crate as a result of the 0.5m\begin{align*}0.5 \;\mathrm{m}\end{align*} movement.
4. Determine the change in gravitational potential energy,  GPE\begin{align*}\triangle\ GPE\end{align*}, of each crate as a result of the 0.5m\begin{align*}0.5 \;\mathrm{m}\end{align*} movement.

### The Inclined Plane

The inclined plane, a simple machine devised by ancient man, is used even today as a routine way to help move bodies more easily to a higher level. The sloping surfaces at loading docks and the wheelchair ramps outside hospitals and schools are just a few applications of the inclined plane. In this section, we will apply several ideas of forces and work involving a frictional force to the inclined plane.

It takes more effort and force to lift a body up a staircase or ladder than to simply push a body up a frictionless inclined plane. The inclined plane therefore has a mechanical advantage. When comparing the length of a staircase or ladder to the length of an inclined plane going from the same height, h\begin{align*}h\end{align*}, to ground level, the length of the staircase or ladder is much shorter. The trade-off for putting in less effort and force in pushing a body up an incline is increased displacement. As we explained earlier, the amount of work done by gravity to lower a body form a certain height, h\begin{align*}h\end{align*}, to ground level is independent of the path because gravity is a conservative force. The amount of work done against gravity to raise a body from ground level to height, h\begin{align*}h\end{align*}, is, therefore, also independent of the path.

Figure 13 shows a crate of mass, m=2.0kg\begin{align*}m = 2.0 \;\mathrm{kg}\end{align*}, placed on a ramp inclined at an angle of 30o\begin{align*}30^o\end{align*} to the horizontal ground. The coefficients of static friction and kinetic friction are μs=0.4\begin{align*}\mu_s = 0.4\end{align*} and μk=0.2\begin{align*}\mu_k = 0.2\end{align*}, respectively, between the crate and the incline. The straight-line distance from the crate to the end of the incline is 2.0m\begin{align*}2.0 \;\mathrm{m}\end{align*}. Note: A motion sensor, placed at the bottom of the incline, detects the instantaneous velocity of crate 1\begin{align*}1\end{align*}, which can be used to determine the kinetic energy, KE, of the crate.

This illustrates a crate placed on a ramp that is inclined to the horizontal. The coefficients of static friction and kinetic friction are and , respectively, between the crate and the incline. The straight-line distance from the crate to the end of the incline is .

Procedural Steps

1. In the boxes to the right of the mass-pulley system in Figure 13, draw a free-body diagram of the crate, showing all the forces on it. Label each force and its magnitude.
2. Determine the work done by each force on the crate and the total work, WT\begin{align*}W_T\end{align*}, done on the crate.
3. Using the work-energy theorem, determine the change in kinetic energy,  KE\begin{align*}\triangle\ KE\end{align*}, of the crate as a result of the 2.0m\begin{align*}2.0 \;\mathrm{m}\end{align*} movement.
4. Determine the change in gravitational potential energy,  GPE\begin{align*}\triangle\ GPE\end{align*}, of the crate as a result of the 2.0m\begin{align*}2.0 \;\mathrm{m}\end{align*} movement down the incline. Is gravity a “conservative” force? (i.e., work done is independent of the path taken)

## Review Questions

1. Starting from ground level, a pulling force, F=200lb\begin{align*}F = 200 \;\mathrm{lb}\end{align*} pulls a 500lb\begin{align*}500 \;\mathrm{lb}\end{align*} crate for 40ft\begin{align*}40 \;\mathrm{ft}\end{align*} up a ramp at constant speed. The ramp makes an angle of 20o\begin{align*}20^o\end{align*} with the horizontal ground. The vertical rise is 11.7ft\begin{align*}11.7 \;\mathrm{ft}\end{align*}. 1.Calculate the work done by F\begin{align*}F\end{align*} alone, WF\begin{align*}W_F\end{align*}. 2.Calculate work done by gravity, Wg\begin{align*}W_g\end{align*}. 3.Calculate the work done by kinetic friction, Wf\begin{align*}W_f\end{align*}, in pulling the crate 40ft\begin{align*}40 \;\mathrm{ft}\end{align*} up the inclined plane. 4.Calculate the ΔGPE\begin{align*}\Delta GPE\end{align*} of crate when it reaches the 11.7ft\begin{align*}11.7 \;\mathrm{ft}\end{align*} vertical height.

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